\begin{tabular}{|l|l|l|l|}\hline \( \begin{array}{l}\text { A deck contains } 8 \text { cards labeled } 1 \text { through } 8 \text {. A card is randomly } \\ \text { selected and returned to the deck. The deck is shuffled. Then } \\ \text { another random selection is made. }\end{array} \) & \( \begin{array}{l}\text { Event A: The first } \\ \text { selection is a } 3 .\end{array} \) \\ \hline & \( \begin{array}{l}\text { Event B: The second } \\ \text { selection is a } 6 .\end{array} \) \\ \hline A family has two children. & \( \begin{array}{l}\text { Event A: The older } \\ \text { child is a girl. }\end{array} \) \\ \hline Event B: Both \\ children are girls. \\ \hline A number cube with sides labeled 1 through 6 is rolled twice. & \( \begin{array}{l}\text { Event A: The first roll } \\ \text { Is a } 4 .\end{array} \) \\ \hline Event B: The second \\ roll is a 5. & \\ \hline\end{tabular}

Let's analyze the scenarios presented in the table step by step, focusing on the events and their probabilities. ### Scenario 1: Deck of Cards 1. **Event A**: The first selection is a 3. 2. **Event B**: The second selection is a 6. **Calculating Probabilities:** - The probability of selecting a 3 on the first draw (Event A) is: \[ P(A) = \frac{1}{8} \] - The probability of selecting a 6 on the second draw (Event B) is: \[ P(B) = \frac{1}{8} \] Since the selections are independent (the card is returned to the deck), the joint probability of both events occurring is: \[ P(A \cap B) = P(A) \times P(B) = \frac{1}{8} \times \frac{1}{8} = \frac{1}{64} \] ### Scenario 2: Family with Two Children 1. **Event A**: The older child is a girl. 2. **Event B**: Both children are girls. **Calculating Probabilities:** - The possible combinations of two children (assuming equal probability for boys and girls) are: - Boy-Boy (BB) - Boy-Girl (BG) - Girl-Boy (GB) - Girl-Girl (GG) Given that the older child is a girl (Event A), the possible combinations are reduced to: - Girl-Girl (GG) - Girl-Boy (GB) Thus, the probability of Event A occurring is: \[ P(A) = \frac{2}{4} = \frac{1}{2} \] Now, given that the older child is a girl, the probability that both children are girls (Event B) is: \[ P(B | A) = \frac{1}{2} \] ### Scenario 3: Rolling a Number Cube 1. **Event A**: The first roll is a 4. 2. **Event B**: The second roll is a 5. **Calculating Probabilities:** - The probability of rolling a 4 on the first roll (Event A) is: \[ P(A) = \frac{1}{6} \] - The probability of rolling a 5 on the second roll (Event B) is: \[ P(B) = \frac{1}{6} \] Since the rolls are independent, the joint probability of both events occurring is: \[ P(A \cap B) = P(A) \times P(B) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36} \] ### Summary of Results - For the deck of cards: \( P(A \cap B) = \frac{1}{64} \) - For the family with two children: \( P(B | A) = \frac{1}{2} \) - For the number cube: \( P(A \cap B) = \frac{1}{36} \) These calculations provide a clear understanding of the probabilities associated with each scenario and the relationships between the events.